Solve $\sin a =\sin x+\sin y, \cos a =\cos x+\cos y$ for $x$ and $y$ in terms of $a$

Question

I saw this on quora.

Solve

$\begin{array}\\ \sin a &=\sin x+\sin y\\ \cos a &=\cos x+\cos y\\ \end{array} $

for $x$ and $y$ in terms of $a$.

Here is my solution.

My question: is there a better way of solving this?

My answer is $y =a+\frac{\pi}{3} $ and $x =a-\frac{\pi}{3} $.

Squaring and adding, $1 = 2+2\sin(x)\sin(y)+2\cos(x)\cos(y) $ or $\cos(x-y) =-\frac12 $ so $x-y =\pm\frac{2\pi}{3} $ so either $x =y+\frac{2\pi}{3} $ or $x =y-\frac{2\pi}{3} $.

(I'm not worrying about the $2 k\pi$ for now.)

Substituting $x =y+\frac{2\pi}{3} $ in the first,

$\begin{array}\\ \sin a &=\sin x+\sin y\\ &=\sin(y+\frac{2\pi}{3})+\sin y\\ &=\sin(y)\cos(\frac{2\pi}{3})+\cos(y)\sin(\frac{2\pi}{3})+\sin y\\ &=-\frac12\sin(y)+\frac{\sqrt{3}}{2}\cos(y)+\sin y\\ &=\frac12\sin(y)+\frac{\sqrt{3}}{2}\cos(y)\\ &=\sin(\frac{\pi}{6})\sin(y)+\cos(\frac{\pi}{6})\cos(y)\\ &=\cos(y-\frac{\pi}{6})\\ &=\sin(\frac{\pi}{2}-(y-\frac{\pi}{6})) \qquad\text{(using } \cos(z)=\sin(\frac{\pi}{2}-z))\\ &=\sin(\frac{2\pi}{3}-y)\\ \end{array} $

so $y =-a+\frac{2\pi}{3} $ and $x =y+\frac{2\pi}{3} =-a+\frac{4\pi}{3} $.

Checking,

$\begin{array}\\ \sin x+\sin y =\sin(a)\\ \cos x+\cos y =-\cos(a)\\ \end{array} $

so this doesn't work!!!

Note: the checking was done by Wolfram Alpha.

Try the other solution.

Substituting $x =y-\frac{2\pi}{3} $ in the first,

$\begin{array}\\ \sin a &=\sin x+\sin y\\ &=\sin(y-\frac{2\pi}{3})+\sin y\\ &=\sin(y)\cos(-\frac{2\pi}{3})+\cos(y)\sin(-\frac{2\pi}{3})+\sin y\\ &=-\frac12\sin(y)-\frac{\sqrt{3}}{2}\cos(y)+\sin y\\ &=\frac12\sin(y)-\frac{\sqrt{3}}{2}\cos(y)\\ &=\sin(\frac{\pi}{6})\sin(y)-\cos(\frac{\pi}{6})\cos(y)\\ &=-\cos(y+\frac{\pi}{6})\\ &=-\sin(\frac{\pi}{2}-(y+\frac{\pi}{6}))\\ &=-\sin(\frac{\pi}{3}-y)\\ &=\sin(y-\frac{\pi}{3})\\ \end{array} $

so $y =a+\frac{\pi}{3} $ and $x =y-\frac{2\pi}{3} =a-\frac{\pi}{3} $.

Upon checking, this works.

Since $x$ and $y$ play symmetric roles in the equations and your answers are not symmetric, you're definitely missing something. — jjagmath, Mar 27 '23 at 04:15
Related: Vertices of equilateral triangle inscribed in the unit circle with $z_1=\operatorname{cis}(x)$, $z_2=\operatorname{cis}(y)$, $z_3=\operatorname{cis}(a+\pi)$. — dxiv, Mar 27 '23 at 04:23
Related: https://math.stackexchange.com/a/1912849/42969. The points $\cos(x) + i \sin(x)$, $\cos(y) + i \sin(y)$, and $\cos(-a) + i \sin(-a)$ form an equilateral triangle with vertices on the unit circle. — Martin R, Mar 27 '23 at 04:30

score 1 · Answer 1 · answered Mar 27 '23 at 04:26

HINT

I would recommend you to divide both equations in order to get: \begin{align*} \tan(a) = \frac{\sin(x) + \sin(y)}{\cos(x) + \cos(y)} & = \frac{2\sin\left(\frac{x + y}{2}\right)\cos\left(\frac{x - y}{2}\right)}{2\cos\left(\frac{x + y}{2}\right)\cos\left(\frac{x - y}{2}\right)} = \tan\left(\frac{x + y}{2}\right) \end{align*}

Based on it, can you proceed from here?

Solve $\sin a =\sin x+\sin y, \cos a =\cos x+\cos y$ for $x$ and $y$ in terms of $a$

1 Answers1